elementary number theory - Show that if $gcd(abc,d^2)=1$, then $gcd(a,d)=gcd(b,d)=gcd(c,d)=1$.

Let $a,b,c$ be integers. Show that if $\gcd(abc,d^2)=1$, then $\gcd(a,d)=\gcd(b,d)=\gcd(c,d)=1$.

Here is my way of approaching this question:

Suppose $\gcd(abc,d^2)=1$, there exist integers $x,y$ such that $abcx+d^2y=1$

$a(bcx)+d(dy)=1$, which implies that $\gcd(a,d)=1$

$b(acx)+d(dy)=1$, which implies that $\gcd(b,d)=1$

$c(abx)+d(dy)=1$, which implies that $\gcd(c,d)=1$

So far I don't really know if this is the way to answer this question. Any help would be appreciated.